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Question Number 140030 by mathdanisur last updated on 03/May/21

	Answered by mr W last updated on 03/May/21

	Commented by mathdanisur last updated on 03/May/21

	$c o o l t h a n k s s i r$
	Commented by mr W last updated on 03/May/21
	$EG=1×tan 27° BD=(DG+1)tan 21° (EG−BD)tan 39°=DG (tan 27°−tan 21°−DG tan 21°)tan 39°=DG (tan 27°−tan 21°)tan 39°=(1+tan 21°tan 39°)DG DG=(((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°)) FG=1×tan 9° CD=(1+(((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°)))tan 9° AD=BD×tan 39°= =((((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°))+1)tan 21°tan 39° tan ∠CAD=((CD)/(AD))=(((1+(((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°)))tan 9°)/(((((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°))+1)tan 21°tan 39°)) =((tan 9°)/(tan 21°tan 39°)) tan ∠AFG=((FG)/(AD+DG)) =((tan 9°)/(((((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°))+1)tan 21°tan 39°+(((tan 27°−tan 21°)tan 39°)/(1+tan 21°tan 39°)))) =((tan 9°)/(tan 21°tan 39°+(tan 27°−tan 21°)tan 39°)) =((tan 9°)/(tan 27°tan 39°)) tan x=tan (∠ACD−∠AFG) =((((tan 9°)/(tan 21°tan 39°))−((tan 9°)/(tan 27°tan 39°)))/(1+((tan 9°)/(tan 21°tan 39°))×((tan 9°)/(tan 27°tan 39°)))) =((((tan 9°)/(tan 39°))(tan 27°−tan 21°))/((((tan 9°)/(tan 39°)))^2 +tan 21°tan 27°)) x=tan^(−1) {((((tan 9°)/(tan 39°))(tan 27°−tan 21°))/((((tan 9°)/(tan 39°)))^2 +tan 21°tan 27°))} =6°$
	$E G = 1 \times \tan 27 °$ $B D = (D G + 1) \tan 21 °$ $(E G - B D) \tan 39 ° = D G$ $(\tan 27 ° - \tan 21 ° - D G \tan 21 °) \tan 39 ° = D G$ $(\tan 27 ° - \tan 21 °) \tan 39 ° = (1 + \tan 21 ° \tan 39 °) D G$ $D G = \frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °}$ $F G = 1 \times \tan 9 °$ $C D = (1 + \frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °}) \tan 9 °$ $A D = B D \times \tan 39 ° =$ $= (\frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °} + 1) \tan 21 ° \tan 39 °$ $\tan ∠ C A D = \frac{C D}{A D} = \frac{(1 + \frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °}) \tan 9 °}{(\frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °} + 1) \tan 21 ° \tan 39 °}$ $= \frac{\tan 9 °}{\tan 21 ° \tan 39 °}$ $\tan ∠ A F G = \frac{F G}{A D + D G}$ $= \frac{\tan 9 °}{(\frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °} + 1) \tan 21 ° \tan 39 ° + \frac{(\tan 27 ° - \tan 21 °) \tan 39 °}{1 + \tan 21 ° \tan 39 °}}$ $= \frac{\tan 9 °}{\tan 21 ° \tan 39 ° + (\tan 27 ° - \tan 21 °) \tan 39 °}$ $= \frac{\tan 9 °}{\tan 27 ° \tan 39 °}$ $\tan x = \tan (∠ A C D - ∠ A F G)$ $= \frac{\frac{\tan 9 °}{\tan 21 ° \tan 39 °} - \frac{\tan 9 °}{\tan 27 ° \tan 39 °}}{1 + \frac{\tan 9 °}{\tan 21 ° \tan 39 °} \times \frac{\tan 9 °}{\tan 27 ° \tan 39 °}}$ $= \frac{\frac{\tan 9 °}{\tan 39 °} (\tan 27 ° - \tan 21 °)}{{(\frac{\tan 9 °}{\tan 39 °})}^{2} + \tan 21 ° \tan 27 °}$ $x = \tan^{- 1} {\frac{\frac{\tan 9 °}{\tan 39 °} (\tan 27 ° - \tan 21 °)}{{(\frac{\tan 9 °}{\tan 39 °})}^{2} + \tan 21 ° \tan 27 °}}$ $= 6 °$
	Commented by Tawa11 last updated on 23/Jul/21

	$great$
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